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Triangle problem

  1. Dec 23, 2007 #1
    1. The problem statement, all variables and given/known data
    A point P is located in the interior of an equilateral triangle ABC. Perpendiculars drawn from P meet each of the sides in points D,E,F, respectively. Where should P be located to make PD + PE + PF a maximum? What about a minimum?


    2. Relevant equations



    3. The attempt at a solution

    From symmetry considerations. I think that the maximum will be at the incenter. I am not sure about the minimum. Maybe at a vertex, but that is not really in the interior... I tried the standard reflections. I am not sure if I actually need to set up a coordinate system with P at the origin... That looks like it could get messy? I am also not sure how to use the fact that this is an equilateral triangle.
     
  2. jcsd
  3. Dec 23, 2007 #2
    anyone?
     
  4. Dec 24, 2007 #3
    Isn't this one of them Lagrange multiplier things ?

    Having said that, i don't know how to set up :

    1)the funtion that needs to be minimized/maximized

    2) the function that is going to be the constraint.

    I would reallly like to see the solution to this problem, though.

    Marlon
     
  5. Dec 24, 2007 #4
    I think it's my english that's bad here, but how can you draw 3 perpendicular lines in a plane?
     
  6. Dec 24, 2007 #5
    They are not perpendicular to each other.

    They are perpendicular to each side of the triangle.
     
  7. Dec 24, 2007 #6
    How about this approach: (sorry for the double post)

    Draw segments from P to each vertex of the triangle. You have split the triangle into three new triangles.

    The area of each triangle is given by [tex]\frac{1}{2}sh_i[/tex] where s is the side length of the equilateral triangle and h_i is the height of the perpendicular on that side

    Through this, we can set up the equality

    [tex]\frac{1}{2}sh_a + \frac{1}{2}sh_b + \frac{1}{2}sh_c = \frac{s^2 \sqrt{3}}{4}[/tex]

    [tex]h_a + h_b + h_c = \frac{s \sqrt{3}}{2}[/tex]

    meaning that the sum of the three perpendiculars is always constant regardless of the location of P
     
  8. Dec 24, 2007 #7
    well, take the situation that P is at point a, then if you move D a distance d, for instance, you can still hold E still, but you'll have to move F to keep P one point.

    now P moves in the direction of E with a distance: 2d/root[3], this is the distance it shrinks.
    the distance it grows is: 2d tan(30) = 2d/root[3], this is equal.

    because any motion of P can be described as the sum of the motion in the direction of perpendiculars, the distance is always the same.
     
    Last edited: Dec 24, 2007
  9. Dec 24, 2007 #8
    I must have done something wrong, because chickendude seems right.

    EDIT: I've edited my post and saw what i did wrong.
     
    Last edited: Dec 24, 2007
  10. Dec 24, 2007 #9
     
  11. Dec 24, 2007 #10
    Hmmm. I did not see anything wrong with chickendude's post, which means there is probably something wrong here.
     
  12. Dec 24, 2007 #11

    NateTG

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    Homework Helper

    If you consider the point [itex]p[/itex] to be the tip of three triangles, each of which has one of the sides of the original equilateral triangles as it's base then the total area of those triangles will be:
    [tex]\frac{1}{2} l_1 s + \frac{1}{2} l_2 s + \frac{1}{2} l_3 s = \frac{1}{2} (l_1 + l_2 +l_3) s[/tex]
    which is also the area of the original triangle, which is clearly constant, so
    [tex](l_1+l_2+l_3)[/tex]
    must be constant.
     
  13. Dec 24, 2007 #12
    Yes. You just repeated chickendude's proof. I am just saying if that is right, then something is wrong with Rainbow Child's proof.
     
  14. Dec 24, 2007 #13
    My mistake!

    In my proof I took the points

    [tex]E(\frac{\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\,F (\frac{3\,\alpha}{4},\frac{\sqrt{3}\,\alpha}{4}),\ ,D(\frac{\alpha}{2},0) [/tex]

    which are the midpoints of the sides AB, AC and AD! (nothing to do with the original problem :smile:)

    So I gave the answer to the question:

    "Which point extemizes the sum [tex] |PD|+|PE|+|PF| [/tex] when D, E and F are the midpoints of the sides!"

    Sorry!
     
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